 Hi, and welcome to your session. Here is this cast of calling question. The question says, integrate the following functions, x cubed by square root of 1 minus x to the power 8. Let's now begin with the solution. In this question, we have to integrate the function, x cubed by square root of 1 minus x to the power 8 with respect to x. Now, this is equal to integral of x cubed by square root of 1 minus. We can write x to the power 8 as x to the power 4 whole square tx. Now, we will put x to the power 4 as t. Now, this implies 4x cubed is equal to dt by dx. This implies x cubed dx is equal to dt by 4. So it's x to the power 4 by p and x cubed dx by dt by 4. So now, this integral is equal to 1 by 4 into integral of dt by square root of 1 minus t square. We know that integral of 1 by square root of a square minus x square dx is sine inverse x by a plus c. Now here, in place of x, we have t and in place of a, we have 1. So integral of dt by square root of 1 minus t square is equal to 1 by 4 into sine inverse t by 1 plus c. We have substituted t in place of x to the power 4. So this is equal to 1 by 4 sine inverse x to the power 4 plus c. Hence, our required answer is 1 by 4 sine inverse x to the power 4 plus c. So this completes the session. Bye and take care.